Explanation

(a)
$\sum_{r = 0}^{m}{\ ^{n + r}C_{n}} = \sum_{r = 0}^{m}{\ ^{n + r}C_{r}}$
$= \ ^{n}C_{0} + \ ^{n + 1}C_{1} + \ ^{n + 2}C_{2} + ... + \ ^{n + m}C_{m}$
$= \ ^{n + 1}C_{0} + \ ^{n + 1}C_{1} + \ ^{n + 2}C_{2} + ... + \ ^{n + m}C_{m}$
$\lbrack\because\ ^{n + 1}C_{0} = \ ^{n}C_{0}\rbrack$
$= \ ^{n + 2}C_{1} + \ ^{n + 2}C_{2} + ... + \ ^{n + m}C_{m}$
$= \ ^{n + m}C_{m - 1} + \ ^{n + m}C_{m}$
$= \ ^{n + m + 1}C_{m}$ $\left\lbrack \because\ ^{n}C_{r - 1} + \ ^{n}C_{r} = \ ^{n + r}C_{r} \right\rbrack$
$= \ ^{n + m + 1}C_{n + 1}$ [$\because\ ^{n}C_{r} = \ ^{n}C_{n - r}$]